Find the area of the region inside the circle (x-1)^2+y^2=1 and outside the circle x^2+y^2=1

There are numerous ways to approach this. Perhaps the simplest is to subtract twice the area of the shaded segment (see the attached figure) from that of the circle.

The area of the shaded segment is
  segment area = (1/2)r^2*(θ -sin(θ))
where r=1 and θ=2π/3

The area of the circle, of course, is πr^2 = π, so the area of interest is ...
  A = π -(2π/3 -sin(2π/3))
  A = π/3 +(√3)/2

Step-by-step explanation:

4). m(arc GJ) = 90°

5). m(arc HI) = m(arc JH) - m(arc JI)

                    = 180° - 151°

                    = 29°

6). m(arc HIJ) = 180°

7). m(arc GJI) = m(arc GJ) + m(arc JI)

                      = 90° + 151°

                      = 241°

8). m(arc GHJ) = m(arc GH) + m(arc HI) + m(arc IJ)

                        = 90° + 29° + 151°

                        = 170°

9). m(arc GJH) = 360° - m(arc GH)

                        = 360° - 90°

                        = 270°

10). m(arc HGJ) = 180°

11). m(arc GH) = 90°

12). m(arc GHI) = m(arc GH) + m(arc HI)

                        = 90° + 29°

                        = 119°

13). m(arc HJI) = 360° - m(arc HI)  

                       = 360° - 29°

                       = 331°

14). m(arc JHI) = m(arc JH) + m(arc HI)

                       = 180° + 29°

                       = 209°

15). m(arc JIG) = 360° - m(arc JG)

                       = 360° - 90°

                       = 270°